 New Zealand
Level 8 - NCEA Level 3

Applications of Parametric Equations

Lesson

Many applications of parametric equations concern problems of motion. The parameter in such cases is usually time $t$t and the equations describe the motion in mutually perpendicular directions measured by the $x$x- and $y$y-axes.

We could have an equation for motion in the $z$z-direction as well but for the time being we will confine our attention to two-dimensional motion only.

Example 1

An object is moving according to the parametric equations

$x=2t^2$x=2t2
$y=-9t+36\sqrt{t}$y=9t+36t

From the second equation, we see that when $t=0$t=0 we have $y=0$y=0. Is there another value of $t$t that makes $y=0$y=0? (This would correspond to a projectile launched from the ground at $t=0$t=0 returning to the ground sometime later.)

We need to solve $0=-9t+36\sqrt{t}$0=9t+36t. One way to do this is to factorise the right-hand side to obtain $0=\sqrt{t}\left(-9\sqrt{t}+36\right)$0=t(9t+36)

A solution to this equation is $\sqrt{t}=0$t=0 and hence, $t=0$t=0, which we already knew. The other solution, if there is one, comes from setting the other factor equal to zero.

 $0$0 $=$= $-9\sqrt{t}+36$−9√t+36 $=$= $\sqrt{t}-4$√t−4 $\therefore\ \ \ \sqrt{t}$∴   √t $=$= $4$4 $t$t $=$= $16$16

What distance in the $x$x-direction is covered in the interval between $t=0$t=0 and $t=16$t=16? That is, we want to find the range of the object.

Using the second parametric equation, we find that when $t=0$t=0, $x=0$x=0 and when $t=16$t=16, $x=2\times16^2=512$x=2×162=512. Therefore the distance is $512$512 units.

What is the rectangular equation of the shape of the path of the object?

We need to eliminate $t$t from the parametric equations. From the first equation, we deduce two possible expressions for $t$t: $t=\pm\sqrt{\frac{x}{2}}$t=±x2. However, since in the second parametric equation we will require $\sqrt{t}$t to be a real number, we will use the positive value $t=\sqrt{\frac{x}{2}}$t=x2.

Therefore, $y=-9\sqrt{\frac{x}{2}}+36\sqrt{\sqrt{\frac{x}{2}}}$y=9x2+36x2.

The graph of this equations is as follows: We can find the maximum $y$y-value attained by the moving object from the second of the parametric equations.

We have $y=-9t+36\sqrt{t}$y=9t+36t. This is a quadratic in $\sqrt{t}$t which we can write more conveniently as $y=-9a^2+36a$y=9a2+36a where we have put $\sqrt{t}=a$t=a.

This is

 $y$y $=$= $-9\left(a^2-4a+4-4\right)$−9(a2−4a+4−4) $=$= $-9\left((a-2)^2-4\right)$−9((a−2)2−4)

This expression will be maximised when the $(a-2)^2$(a2)2 term is minimised. That is, when $a=2$a=2 or $t=4$t=4 and so, $y=36$y=36. This can be checked from the graph.

Example 2

An object is shown moving across a radar screen so that in $1$1 minute it has travelled $40$40 km towards the east and $50$50 km towards the north.

a. Find the velocities in km/h in the two directions.

$40$40 km in $1$1 minute is $40\times60=2400$40×60=2400 km/h
$50$50 km in $1$1 minute is $50\times60=3000$50×60=3000 km/h

b. Write a pair of parametric equations giving distances as functions of time.

We put $x$xas the distance towards the east and $y$y as the direction towards the north. Thus $x=40t$x=40t and $y=50t$y=50t with time in minutes.

c. What absolute distance is traversed in $2$2 minutes?

In $2$2 minutes, $x=40\times2=80$x=40×2=80 km and $y=50\times2=100$y=50×2=100 km. So, we calculate $\sqrt{80^2+100^2}=128$802+1002=128 km.

d. Give the trajectory as a relation between distances in the two directions.

From $x=40t$x=40t and $y=50t$y=50t, we obtain $t=\frac{x}{40}$t=x40 and therefore, $y=50\times\frac{x}{40}$y=50×x40. Thus, the trajectory is given by $y=\frac{5x}{4}$y=5x4.

Worked Examples

Question 1

The position in metres of a projectile $t$t seconds after its launch is described by $x=55t$x=55t and $y=102t-17t^2$y=102t17t2.

1. Find the time $t$t at which the projectile returns to the ground.

2. Find the maximum horizontal distance covered by the projectile.

3. Find the maximum height reached by the projectile.

4. Find the rectangular equation for the path of the projectile.

Question 2

A rock is thrown from the top of a $150$150 m tower with an initial speed of $28$28 m/s at an angle of $60^\circ$60°.

1. Find $v_x$vx, the horizontal component of the initial velocity. Give your answer in exact form with a rational denominator.

2. Find $v_y$vy, the vertical component of the initial velocity. Give your answer in exact form with a rational denominator.

3. State the parametric equation $x$x that models the horizontal component of the path of the rock.

4. The vertical displacement in metres of a projectile is given by the expression $h+vt\sin\theta-5t^2$h+vtsinθ5t2, where $h$h is the initial height, $v$v is the initial velocity and $\theta$θ is the angle of elevation.

Use this information and your previous answers to state the parametric equation $y$y that models the vertical component of the path of the rock $t$t seconds after it is thrown.

5. Find the number of seconds $t$t that the rock was in the air.

6. How far from the base of the tower does the rock land?

Outcomes

M8-1

Apply the geometry of conic sections

91573

Apply the geometry of conic sections in solving problems